A very simple question: what spaces of function does the Laplace transform map from and into? Given a function $f$, we can write $f\colon\mathbb{R} \to \mathbb{R}$ to denote that $f$ takes a number from $\mathbb{R}$ into $\mathbb{R}$. Easy enough.
Given the Laplace transform operator $(\mathcal{L}f)(s) = \int_{0}^{\infty} e^{-st} f(t) dt$, can we in a similar fashion write: $(\mathcal{L}f)(s)\colon A \to B$ where $A,B$ are function spaces?
What are $A$ and $B$ precisely speaking?
What about the Fourier transform?  $\int_{0}^{\infty} e^{-iwt} f(t) dt$
 A: I'm not sure about the Laplace transform but in Joel L. Schiff's "The Laplace Transform: Theory and Applications" on page 13 the author proves that a large class of functions has a Laplace transform. I am not sure how to describe nicely the result in terms of domain and range of the operator, buy maybe that helps.
As for the Fourier transform, you first define it with domain $L^1(\mathbb{R})$, in which case the range (by Riemann-Lebesgue lemma) will be $\mathcal{C}_0(\mathbb{R})$ (i.e. the set of functions that go to zero at infinity). 
Then you can restrict the domain to $L^1\cap L^2$ and notice that this contains the Schwartz space $\mathcal{S}$ which is dense in $L^2(\mathbb{R})$ and so you can extend $\mathcal{F} \colon L^2(\mathbb{R}) \to L^2(\mathbb{R})$ (so domain and range both being $L^2$) which also turns out to be an isometry. 
Moreover, if you restrict your attention to $\mathcal{S}$, you get $\mathcal{F} \colon \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R})$ and similarly $\mathcal{F}\colon \mathcal{S}'(\mathbb{R}) \to \mathcal{S}'(\mathbb{R})$, where $\mathcal{S}'(\mathbb{R})$ is the set of tempered distributions. More things can be said about the Fourier transform on $L^p$, for $p \in (1,2]$ but this is less classical and possibly less interesting.
